This brief video is a review of circuit manipulations using ideal transformers. The basic rules for ideal transformers, in which we push an element, a circuit element through a transformer to, what we, what we called reflecting to the other side goes as follows. If we have a voltage and we push it through a transformer. That has a turns ratio of n1 to n2. [SOUND]. Then, as seen between the secondary terminals of the transformer. Effectively, we have a voltage source whose value is multiplied by the turns ratio. So we get n2 over n1 times v. As the effective voltage source as seen by the second, from the secondary terminals. Likewise with a current source, we can push it through a transformer also. But we have to divide by the turns ratio. So if we have an n1 to n2 turns ratio here Then as seen between the secondary terminals we effectively have a current source of value y that is divided by n2 over n1. Or we can write n1 over n2 times y. And the third manipulation is if we have some impedence. We'll call it z. When we push that through a transformer. And we effectively see an impedence between the secondary terminals. That is the turns ratio squared times z. So we get n 2 over n 1 squared z as the effective impedance seen between the secondary terminals. one little case to be careful with here. With z is if z is a resistor so if we have a resistor then effectively between the secondary terminals. The value will be the resistance times the turns ratio squared. Likewise if we have an inductor whose impedance is SL, then we would multiply that impedance by the same turns ratio squared. [SOUND] Which is as if the inductance was multiplied by n2 over n1 squared. So sometimes we just simplify this, to say there's an effective inductor, a value n2 over n1 squared L. If we have a capacitor. Recall that the capacitor, the impedance of a capacitor is 1 over SC. So the impedance here would be n2 over n1 squared times 1 over SC. As far as the effective value of the capacitence, you can see that since impedence varies inversly with capacitence. This is as if we have a capacitor that get's divided by the turns ration squared, instead of multiplied, so it's as if we have a capacitor value C that is, n1 squared over n2 squared. Okay? Those are the rules for pushing simple elements through transformers. Let's do an example now. Suppose we have, say a voltage. Source, and let's take some larger circuit, maybe with a resistor R1, some transformer with a turns ratio, let's say, one to nA. Maybe we have another resistor here of value R2, and another transformer that has the dots reversed. So, the primary dot is on top and the secondary dot is on the bottom of the winding. And let's give this a turns ratio of nB to 1. And what we would like to do is find the effective network with respect to the output terminals. And, in fact, maybe we'll s-, we'll even solve for the output voltage, V out. So to, one good way to solve this circuit is to use these circuit manipulations and push the elements through the transformers. So what I'm going to do here is push them through the transformers one at a time. So let's start by pushing the voltage source and R1 Through the first transformer. So what we will get with respect to the secondary terminals of this first transformer, we'll have the voltage source pushed through and in the process it gets multiplied by nA. So we have n sub av as the voltage, the effective voltage. And it's in series with an effective resistor. That would be the resistance, r1, multiplied by the turns ratio squared. So, nA squared. That's what we see, effectively, between the secondary terminals of this first transformer, okay? So then I'll copy the rest of the circuit after that. Okay, next let's push the elements through the second transformer. Okay, so here is our, still our output terminals. So we'll push the elements to see how they appear effectively between the secondary terminals of this second transformer. So when we push R2 through the transformer, we'll get a resistor on the secondary side. And it gets multiplied by the turns ratio squared. But here, the turns ratio is 1 over nB. So effectively, we would get R2 over nB squared as the effective resistance. The fact that the dots are reversed doesn't effect impedances. Because we, the thought is like to having a minus sign on the turns ratio. But the impedance changes by the turns ratio squared, and the minus sign cancels out. Okay. Next let's push the R1 resistor through. So will get an effective resistor here that is the nA squared R1 and then we have to divide by nB squared for this impedance also. Okay? And then finally we push the voltage source through the transformer. So we have a voltage of value nAV and then we'll divide by this nB and since the dots are reversed, we have to reverse the polarity of the voltage source. So I'm going to draw it with minus on top and plus on the bottom. Okay, so this is the effective network that is seen between the secondary terminals of this, the last transformer. With this circuit now we can solve. So what is the output voltage? We could just use the divider, voltage divider formula so you would take the voltage source, nA over nB times V. And because the polarity's reversed, we have a minus sign. And then that voltage gets multiplied by the divider ratio of these two resistors. So it would be times the divider ratio R2 over NB squared, all over what? The same R2 over NB squared. Plus the first resistor NA squared over NB squared times R1. So, that is the solution then, for the output voltage. So, we've taken a fairly complex circuit, and by manipulating it, reduced it to something where we can just write the answer in the end. One other thing I would point out is that you do have to be a little bit careful in manipulating here. If we had wanted to solve for some other signal that's not at the output. Like suppose we wanted to find this current right here. I'm going to call that IA. IA does not explicitly appear in this final circuit. And in fact, when we push the elements through the transformers, we have to be careful. On the second circuit, IA is this current. And what happens to IA on the third circuit? Well, when we push these elements through the transformer, we end up pushing IA through the transformer also. So here's, here's a current related to IA, but IA got pushed through this turns ratio. And the current on this side would actually be nB times IA. [COUGH] Okay? So we could solve this circuit for that current, but we would be solving for nBIA. And then it, we have to divide by nB to get back what the actual IA is.